If $a> b> c $ and the system of equations $ax+ by +cz=0, bx + cy + az= 0$ and $cx+ay + bz =0$ has a non-trivial solution, then the quadratic equation $ax^2 + bx + c = 0 $ has

at least positive root

The correct answer is option (1) : at least positive root

The given system of equations has a non-trivial solution

$∴\begin{bmatrix} a& b & c\\ b & c & a\\ c & a & b \end{bmatrix}=0$

$⇒(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0 $

$⇒ (a+b+c) [(a+b)^2 + (b-c)^2 + (c-a)^2]=0$

$⇒a+b+c=0 $            $[∵(a-b)^2 + (b-c)^2 +(c-a)^2 ≠0]$

 $x= 1 $ is a root of $ax^2 + bx + c = 0 $

Thus, the quadratic equation has at least one positive root.